Anyone know sebastian from bb

Construct a link mechanism of crank oa 30mm rotating clockwise rod ab 100mm and bc 50mm

m_a_m_a [10]

Answer:

the answer is 180mm

Explanation:

the answer is 180mm.

5 0

2 years ago

g The parameters of a certain transmission line operating at 휔휔=6 ×108 [rad/s] are 퐿퐿=0.35 [휇휇H/m], 퐶퐶=75 [pF/m], 퐺퐺=75 [휇휇S/m],

yKpoI14uk [10]

Explanation:

$\begin{aligned}\gamma &=\sqrt{Z Y}=\sqrt{(R+j \omega L)(G+j \omega C)} \\&-\sqrt{|17|} j\left(6 \times 10^{8}\right)\left(0.35 \times 10^{-6}\right)|| 75 \times 10^{-6}\left|j\left(6 \times 10^{8}\right)\left(40 \times 10^{-12}\right)\right| \\&=0.094+j 2.25 \mathrm{m}^{-1}-\alpha+j \beta\end{aligned}$

Therefore,

$-\alpha-0.094 \mathrm{Np} / \mathrm{m} . \quad 3-2.25 \mathrm{rad} / \mathrm{m}, \text { and } \lambda-2 \pi / \beta-\underline{2.79} \mathrm{m}$

$Z_{0}-\sqrt{\frac{Z}{Y}}-\sqrt{\frac{R+j \omega L}{G+j \omega C}}-\sqrt{\frac{17+j 2.1 \times 10^{2}}{75 \times 10^{-6}+j 2.4 \times 10^{-2}}}-\frac{93.6-j 3.64 \Omega}{4}$

5 0

3 years ago

An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate is found to be 1% pe

LenaWriter [7]

Answer:

a) Q = 251.758 kJ/mol

b) creep rate is $= 1.751 \times 10^{-5} \% per hr$

Explanation:

we know Arrhenius expression is given as

$\dot \epsilon =Ce^{\frac{-Q}{RT}$

where

Q is activation energy

C is pre- exponential constant

At 700 degree C creep rate is $\dot \epsilon = 5.5\times 10^{-2}$ % per hr

At 800 degree C creep rate is $\dot \epsilon = 1$ % per hr

activation energy for creep is $\frac{\epsilon_{800}}{\epsilon_{700}}$ = $= \frac{C\times e^{\frac{-Q}{R(800+273)}}}{C\times e^{\frac{-Q}{R(700+273)}}}$

$\frac{1\%}{5.5 \times 10^{-2}\%} = e^{[\frac{-Q}{R(800+273)}] -[\frac{-Q}{R(800+273)}]}$

$\frac{0.01}{5.5\times 10^{-4}} = ln [e^{\frac{Q}{8.314}[\frac{1}{1073} - \frac{1}{973}]}]$

solving for Q we get

Q = 251.758 kJ/mol

b) creep rate at 500 degree C

we know

$C = \epsilon e^{\frac{Q}{RT}}$

$=- 1\% e{\frac{251758}{8.314(500+273}} = 1.804 \times 10^{12} \% per hr$

$\epsilon_{500} = C e^{\frac{Q}{RT}}$

$= 1.804 \times 10^{12} e{\frac{251758}{8.314(500+273}}$

$= 1.751 \times 10^{-5} \% per hr$

4 0

2 years ago

PythonA group of statisticians at a local college has asked you to create a set of functionsthat compute the median and mode of

skelet666 [1.2K]

Answer:

def median(l):
if(len(l) == 0):
return 0
else:
l.sort()
if(len(l)%2 == 0):
index = int(len(l)/2)
mid = (l[index-1] + l[index]) / 2
else:
mid = l[len(l)//2]
return mid
def mode(l):
if(len(l)==0):
return 0
mode = max(set(l), key=l.count)
return mode
def mean(l):
if(len(l)==0):
return 0
sum = 0
for x in l:
sum += x
mean = sum / len(l)
return mean
lst = [5, 7, 10, 11, 12, 12, 13, 15, 25, 30, 45, 61]
print(mean(lst))
print(median(lst))
print(mode(lst))

Explanation:

Firstly, we create a median function (Line 1). This function will check if the the length of list is zero and also if it is an even number. If the length is zero (empty list), it return zero (Line 2-3). If it is an even number, it will calculate the median by summing up two middle index values and divide them by two (Line 6-8). Or if the length is an odd, it will simply take the middle index value and return it as output (Line 9-10).

In mode function, after checking the length of list, we use the max function to estimate the maximum count of the item in list (Line 17) and use it as mode.

In mean function, after checking the length of list, we create a sum variable and then use a loop to add the item of list to sum (Line 23-25). After the loop, divide sum by the length of list to get the mean (Line 26).

In the main program, we test the three functions using a sample list and we shall get

20.5

12.5

12

3 0

3 years ago

6.Identification of Material ParametersThe principal in-plane stresses and associated strains in a plate of material areσ1= 50 k

ad-work [718]

Answer:

See attached image for diagrams and solution

6 0

2 years ago